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%%\author{王立庆（2019级数学与应用数学1班）}
%\author{学号 \underline{\hspace{4cm}}\,\,\,\, 姓名 \underline{\hspace{4cm}}  }
%%\title{高等代数第六章：向量空间}
%\title{统计软件考试解答 }
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{\Large\bf \H 上海立信会计金融学院期终考试卷 } \hspace{0.3cm} {\Large \underline{ A }卷 解答}

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{\large \bf \H 2022 $\sim$ 2023 学年 第 二 学期 }

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{\large \bf \H \underline{ \emph{2021级数学与应用数学专业} } 《\underline{ \emph{统计软件)} }》 课程代码：\underline{ 160960214 }  }

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本次考试主要是对概率统计问题和R语言代码的阅读理解。请用完整的句子：
\begin{enumerate}\itemsep0em
\item[(a)]  解释题目里提出的问题并回答。 
\item[(b)]  解释带\#\#的代码的含义（写在代码的旁边）。
\item[(c)]  写下对本题的进一步思考。
\end{enumerate}

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\begin{enumerate}

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\item %1 mss - page 85 - 例 2.4.3 
（共12分）甲、乙两棋手进行 20 局比赛，以赢的局数多者为胜。
设在每局中甲赢的概率为 0.52, 乙赢的概率为 0.48. 
设各局比赛是独立进行的。
使用R语言辅助计算，求甲胜、乙胜、不分胜负的各个概率。

\begin{lstlisting}[language=R]
> k=11:20  
> x=choose(20,k)*(0.52)^k*(0.48)^(20-k)  ##(1) 
> sum(x)  ##(2) 
[1] 0.4834428
> k=0:9
> y=choose(20,k)*(0.52)^k*(0.48)^(20-k)
> sum(y)
[1] 0.3431591
> k=10
> z=choose(20,k)*(0.52)^k*(0.48)^(20-k)
> z
[1] 0.1733981

> sum(dbinom(11:20,size=20,prob=0.52)) ##(3)
[1] 0.4834428
> sum(dbinom(0:9,size=20,prob=0.52))
[1] 0.3431591
> dbinom(10,size=20,prob=0.52)
[1] 0.1733981
\end{lstlisting}

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\vspace{0.1cm}

{\color{red}解答：

\begin{enumerate}
\item  %(1)
记随机变量 $X$ 为甲赢的局数。则 $X$ 服从二项分布 $b(n,p)$, 其中 $n=20$, $p=0.52$. 
因此甲胜、乙胜、不分胜负的概率分别是
\begin{align*}
P(X\ge 11) &= \sum\limits_{k=11}^{20} \binom{20}{k} (0.52)^k(0.48)^{20-k} = 0.4834. \\ 
P(X\le 9) &= \sum\limits_{k=0}^{9} \binom{20}{k} (0.52)^k(0.48)^{20-k} = 0.3432. \\ 
P(X= 10) &= \binom{20}{10} (0.52)^{10}(0.48)^{10} = 0.1734.  
\end{align*}

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\item %(2)
代码解释：\\ 
(1)使用R语言中的向量化运算，计算二项分布的概率。\\ 
(2)计算这些概率的和。\\
(3)直接使用R语言中的二项分布的概率密度函数 \,{\texttt{dbinom}}, 计算二项分布的概率的和。

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %2 mss - page 16 - 例子1.2.3. 
（共12分）一批产品共有 $N=200$ 件，其中 $M=10$ 件是不合格品， $N-M=190$ 件是合格品。
从中随机取出 $n=20$ 件。
使用R语言辅助计算，求事件 $A_2 = $ “取出的 $n=20$ 件产品中有 $m=2$ 件不合格品”的概率。

\begin{lstlisting}[language=R]
> choose(190,18)*choose(10,2)/choose(200,20)  ##(1) 
[1] 0.1975432
\end{lstlisting}

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{\color{red}解答：
\begin{enumerate}
\item  %(1)
从 $N$ 件产品中不放回地选取 $n$ 件共有 $\binom{N}{n}$ 中可能。事件 $A_m$ 可以分成两步，第一步从 $N-M$ 件合格品中选取 $n-m$ 件，第二步从 $M$ 件不合格品中选取 $m$ 件。所求概率的理论值为
\begin{eqnarray*}
P(A_2) = \frac{\binom{N-M}{n-m}\binom{M}{m}}{\binom{N}{n}} = \frac{\binom{190}{18}\binom{10}{2}}{\binom{200}{20}} 
= 0.1975. 
\end{eqnarray*}

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\item %(2)
代码解释：\\ 
使用组合数计算概率。
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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %3
（共12分）设二维随机变量 $(X,Y)$ 服从二元正态分布，即 $(X,Y)\sim N(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho)$. 
\begin{enumerate}
\item[(1)]  写出 $(X,Y)$ 的联合概率密度函数。
\item[(2)]  使用R语言模拟生成一些服从二元正态分布的随机向量 $(x_k,y_k), 1\le k\le n$. 
\end{enumerate}

\begin{lstlisting}[language=R]
rho = 0.60
n = 500
x = rnorm(n)  ##(1)
y = rnorm(n,rho*x,sqrt(1-rho^2))  ##(2) 
plot(x,y,pch=21, col='blue', xlim=c(-5,5),ylim=c(-5,5))  ##(3) 
\end{lstlisting}

% \begin{figure}[ht]
% \centering
% \includegraphics[height=4cm, width=4cm]{ssr-final-p3.png}
%% \caption{Two dimensional normal distribution. }
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{\color{red}解答：
\begin{enumerate}
\item  %(1)

\begin{enumerate}
\item[(1)]  二维正态分布的联合概率密度函数为
\begin{eqnarray*}
p(x,y)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \exp\left[ -\frac{1}{2(1-\rho^2)} G(x,y) \right], -\infty < x,y < \infty, 
\end{eqnarray*}
其中
\begin{eqnarray*}
G(x,y)=\frac{(x-\mu_1)^2}{\sigma_1^2} - 2\rho \frac{ (x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2} + \frac{(y-\mu_2)^2}{\sigma_2^2}. 
\end{eqnarray*}

\item[(2)]  先在水平方向上生成一维正态分布的随机数，然后对每个随机数，在竖直方向上生成一个正态分布的随机数。
\end{enumerate}
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\item  %(2)
代码解释：\\ 
(1)在 $x$ 轴上生成正态分布的随机数。\\ 
(2)对每个 $x$, 在 $y$ 轴上生成正态分布的随机数。\\ 
(3)画出二元正态分布的模拟数据的散点图。

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %4 mss - page 213 - 例子4.4.3. 正态随机数的产生
（共12分）应用中心极限定理，可以从均匀分布的随机数出发，产生正态分布的随机数。具体方法如下。
\begin{enumerate}
\item[(1)]  设有区间 $[0,1]$ 上的均匀分布的随机数 $x_1,x_2,\cdots,x_{12}$. 
\item[(2)]  计算 $y=x_1+\cdots+x_{12}-6$. 
\item[(3)]  重复(1)-(2), 得到一系列的 $y$. 
\end{enumerate}
使用R语言验证这样得到的随机数 $y$ 近似服从正态分布。

\begin{lstlisting}[language=R]
> n=1000
> x=matrix(runif(12*n),nrow=n)  ##(1) 
> y=rowSums(x)-6  ##(2) 
> hist(y)
> qqnorm(y)  ##(3) 
> shapiro.test(y)  ##(4)

	Shapiro-Wilk normality test

data:  y
W = 0.99866, p-value = 0.6611
\end{lstlisting}

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{\color{red}解答：

\begin{enumerate}
\item  %(1)
使用R语言的 \,{\texttt{runif}} 函数产生均匀分布的随机数，排列成 $n$ 行，每行12个随机数。
使用 \,{\texttt{rowSums}} 函数计算每行的和，并减去6. 最后使用 \,{\texttt{qqnorm}} 和 \,{\texttt{shapiro.test}} 检验是否服从正态分布。

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\item  %(2)
代码解释：\\ 
(1)生成均匀分布的随机数，排列成1000行12列的矩阵。\\ 
(2)每一行的数字加起来，再减去6. \\ 
(3)画出QQ图，检验是否服从正态分布。 \\ 
(4)使用Shapiro检验是否正态分布。

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %5 mss - page 242 - 例子5.3.8.
（共13分）设总体为均匀分布 $U(0,1)$. 设 $x_1,x_2,\cdots,x_n$ 为简单随机样本。设 $x_{(1)},x_{(2)},\cdots,x_{(n)}$ 为次序统计量。
使用R语言验证 $x_{(k)}$ 的数学期望为 $\frac{k}{n+1}$. 

\begin{lstlisting}[language=R]
> N=1000
> n=4
> y=0
> for (m in 1:N){     ##(1) 
+   x=runif(n)        ##(2)
+   y[m]=sort(x)[2]   ##(3)
+ }
> mean(y)  ##(4) 
[1] 0.3949853
> 2/(n+1)
[1] 0.4
\end{lstlisting}

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\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  %(1)
使用 \,{\texttt{runif}} 函数产生均匀分布的随机数。使用 \,{\texttt{sort}} 函数计算次序统计量。这段程序中，设 $ n=4,k=2$, 即均匀分布的四个随机数，从小到大排序的第二个数。

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\item  %(2)
代码解释：\\ 
(1)重复进行1000次试验。\\ 
(2)每次试验都生成在区间$[0,1]$上均匀分布的4个随机数。\\ 
(3)每次试验都取出次序统计量的第2个数。\\ 
(4)计算1000次试验的均值。

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %6 mss - page 305 - 例子6.6.6.
（共13分）设某厂生产的零件的质量服从正态分布 $N(\mu,\sigma^2)$, 单位：克。设从一批产品中抽取10个，测得质量如下。
\begin{center}
\begin{tabular}{cccccccccc} \hline 
18.4 & 21.4 & 17.2 & 18.0 & 21.2 & 22.0 & 20.2 & 21.6 & 19.4 & 20.3 \\ \hline 
\end{tabular}
\end{center}
求总体标准差 $\sigma$ 的0.95 和 0.99 置信区间。

\begin{lstlisting}[language=R]
> x=c(18.4, 21.4, 17.2, 18.0, 21.2, 22.0, 20.2, 21.6, 19.4, 20.3)
> s=sd(x)  ##(1)
> n=length(x)
> s1sq=(n-1)*s^2/qchisq(0.975,n-1)  ##(2)
> s2sq=(n-1)*s^2/qchisq(0.025,n-1)
> s1=sqrt(s1sq)  ##(3)
> s2=sqrt(s2sq)
> s1
[1] 1.142741
> s2
[1] 3.032992

> s1sq=(n-1)*s^2/qchisq(0.995,n-1)  ##(4) 
> s2sq=(n-1)*s^2/qchisq(0.005,n-1)
> s1=sqrt(s1sq)
> s2=sqrt(s2sq)
> s1
[1] 1.026187
> s2
[1] 3.783931
\end{lstlisting}

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{\color{red}解答：
\begin{enumerate}
\item  %(1)
使用卡方统计量 $\chi^2 = \frac{(n-1)S^2}{\sigma^2}$. 在零假设成立时，有 $\chi^2\sim \chi^2(n-1)$. 
根据分位数的定义，可得
$$
\mathbb{P}\left( \chi^2_{\alpha/2} \le \frac{(n-1)S^2}{\sigma^2} \le \chi^2_{1-\alpha/2} \right) = 1- \alpha. 
$$
于是可得总体方差 $\sigma^2$ 的置信区间为
$$
\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \le \hat{\sigma}^2 \le \frac{(n-1)s^2}{\chi^2_{\alpha/2}}. 
$$
因此总体标准差 $\sigma$ 的置信度为 0.95 的置信区间为 $[1.14, 3.03]$, 置信度为 0.99 的置信区间为 $[1.03, 3.78]$.  

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\item  %(2)
代码解释：\\ 
(1)计算数据的标准差。\\ 
(2)按公式计算方差的置信区间的左端，置信度为0.95. \\ 
(3)计算标准差的置信区间的左端。\\ 
(4)按公式计算方差的置信区间的左端，置信度为0.99.

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %7 mss - page 325 - 例子7.2.2.
（共13分）设某产品的长度服从正态分布，其均值设定为 240 毫米。现在从一批产品中抽取10件产品，测得长度如下表。
\begin{center}
\begin{tabular}{cccccccccc} \hline 
242 & 243 & 235 & 245 & 234 & 234 & 237 & 238 & 238 & 253 \\ \hline 
\end{tabular}
\end{center}
判断这批产品的长度是否满足设定要求？$\alpha=0.05$. 

\begin{lstlisting}[language=R]
> x=c(242, 243, 235, 245, 234, 234, 237, 238, 238, 253)
> t.test(x,mu=240)  ##(1)  

	One Sample t-test

data:  x
t = -0.052959, df = 9, p-value = 0.9589
alternative hypothesis: true mean is not equal to 240
95 percent confidence interval:
 235.6284 244.1716
sample estimates:
mean of x 
    239.9 
\end{lstlisting}

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{\color{red}解答：

\begin{enumerate}
\item  %(1)
这是单样本正态总体的均值的检验。因为方差未知，使用 $t$ 统计量进行检验。
使用R语言的 \,{\texttt{t.test}} 函数，结果的 $p$ 值为 $0.9589$. 所以接受零假设，认为这批产品的长度满足设定要求。

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\item  %(2)
代码解释：\\ 
(1) 总体方差未知时对总体均值进行 t 检验。

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\item %8 mss - page 384 - 习题7
（共13分）某粮食加工厂试验三种储藏方法对粮食含水率有无显著影响。
现取—批粮食分成若干份，分别用三种不同的方法（A1, A2, A3）储藏，过一段时间后测得的含水率（单位：\%）如下表。
\begin{center}
\begin{tabular}{c|ccccc} \hline 
A1&7.3&8.3&7.6&8.4&8.3 \\ \hline 
A2&5.4&7.4&7.1&6.8&5.3 \\ \hline 
A3&7.9&9.5&10.0&9.8&8.4 \\ \hline 
\end{tabular}
\end{center}

\begin{enumerate}
\item[(1)]  用R语言的一个数据框来保存这些数据，一列为含水率，另一列为储藏方法。
\item[(2)]  假定各种方法储藏的粮食的含水率服从正态分布，且方差相等。使用方差分析，检验这三种方法对含水率有无显著影响。$\alpha= 0.05$. 
\end{enumerate}

\begin{lstlisting}[language=R]
> x1=c(7.3, 8.3, 7.6, 8.4, 8.3)
> x2=c(5.4, 7.4, 7.1, 6.8, 5.3)
> x3=c(7.9, 9.5, 10.0, 9.8, 8.4)
> x=c(x1,x2,x3)  ##(1)
> A=gl(3,5,15)  ##(2) 
> mydata=data.frame(x=x,A=A)  ##(3) 
> anova(lm(x~A,data=mydata))  ##(4) 
Analysis of Variance Table

Response: x
          Df Sum Sq Mean Sq F value   Pr(>F)    
A          2 18.657  9.3287  13.592 0.000825 ***
Residuals 12  8.236  0.6863                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
\end{lstlisting}


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\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  %(1)

\begin{enumerate}
\item[(1)]  将储藏方法设为因子型变量，用\,{\texttt{gl}} 函数产生。 使用 \,{\texttt{data.frame}} 函数创建数据框。
\item[(2)]  用 \,{\texttt{lm}} 函数进行回归分析，用 \,{\texttt{anova}} 函数进行方差分析。因为程序运行结果 $p$ 值小于 $\alpha$, 所以认为不同储藏方法对粮食的含水率有显著影响。 
\end{enumerate}

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\item  %(2)
代码解释：\\ 
(1)将三行数据排成一行。 \\ 
(2)生成三个水平对应三种方法，每个水平对应五个观测数据。\\ 
(3)将观测数据和相应的水平因子生成数据框。\\ 
(4)将观测数据对不同水平进行回归分析和方差分析。 

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\item  %(3)
思考：看理解给分。
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\end{enumerate}

}

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\end{enumerate}

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